If X~N(μ,σ^2) then what's the distribution for exp(X) ?

  • Thread starter stevenytc
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If X is a r.v. that follows that Gaussian distribution with mean μ and variance σ^2, how do I find the distribution function for exp(X) where exp() is the exponent function.
 

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jbunniii
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If ##Y = \exp(X)##, then the cdf of ##Y## is ##F_Y(\alpha) = P(\exp(X) \leq \alpha)##. As ##\log## is strictly monotonically increasing, this is equivalent to ##P(X \leq \log(\alpha))##, which is ##F_X(\log(\alpha))##, the cdf of ##X## evaluated at ##\log(\alpha)##.

Then you can find the pdf of ##Y## by differentiating ##F_Y(\alpha) = F_X(\log\alpha))## to obtain ##f_Y(\alpha) = f_X(\log(\alpha)) / \alpha##. This is valid for all positive ##\alpha##; if ##\alpha## is negative then the pdf is zero since ##\exp## is never negative.

Proceeding from here should be straightforward as there is a closed form expression for ##f_X##.
 

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